hdu 1944 S-Nim（博弈论-求SG函数）
Arthur and his sister Caroll have been playing a game called Nim for some time now. Nim is played as follows:
The starting position has a number of heaps, all containing some, not necessarily equal, number of beads.
The players take turns chosing a heap and removing a positive number of beads from it.
The first player not able to make a move, loses.
Arthur and Caroll really enjoyed playing this simple game until they recently learned an easy way to always be able to find the best move:
Xor the number of beads in the heaps in the current position (i.e. if we have 2, 4 and 7 the xor-sum will be 1 as 2 xor 4 xor 7 = 1).
If the xor-sum is 0, too bad, you will lose.
Otherwise, move such that the xor-sum becomes 0. This is always possible.
It is quite easy to convince oneself that this works. Consider these facts:
The player that takes the last bead wins.
After the winning player's last move the xor-sum will be 0.
The xor-sum will change after every move.
Which means that if you make sure that the xor-sum always is 0 when you have made your move, your opponent will never be able to win, and, thus, you will win.
Understandibly it is no fun to play a game when both players know how to play perfectly (ignorance is bliss). Fourtunately, Arthur and Caroll soon came up with a similar game, S-Nim, that seemed to solve this problem. Each player is now only allowed to remove a number of beads in some predefined set S, e.g. if we have S =(2, 5) each player is only allowed to remove 2 or 5 beads. Now it is not always possible to make the xor-sum 0 and, thus, the strategy above is useless. Or is it?
your job is to write a program that determines if a position of S-Nim is a losing or a winning position. A position is a winning position if there is at least one move to a losing position. A position is a losing position if there are no moves to a losing position. This means, as expected, that a position with no legal moves is a losing position.
InputInput consists of a number of test cases.
The starting position has a number of heaps, all containing some, not necessarily equal, number of beads.
The players take turns chosing a heap and removing a positive number of beads from it.
The first player not able to make a move, loses.
Arthur and Caroll really enjoyed playing this simple game until they recently learned an easy way to always be able to find the best move:
Xor the number of beads in the heaps in the current position (i.e. if we have 2, 4 and 7 the xor-sum will be 1 as 2 xor 4 xor 7 = 1).
If the xor-sum is 0, too bad, you will lose.
Otherwise, move such that the xor-sum becomes 0. This is always possible.
It is quite easy to convince oneself that this works. Consider these facts:
The player that takes the last bead wins.
After the winning player's last move the xor-sum will be 0.
The xor-sum will change after every move.
Which means that if you make sure that the xor-sum always is 0 when you have made your move, your opponent will never be able to win, and, thus, you will win.
Understandibly it is no fun to play a game when both players know how to play perfectly (ignorance is bliss). Fourtunately, Arthur and Caroll soon came up with a similar game, S-Nim, that seemed to solve this problem. Each player is now only allowed to remove a number of beads in some predefined set S, e.g. if we have S =(2, 5) each player is only allowed to remove 2 or 5 beads. Now it is not always possible to make the xor-sum 0 and, thus, the strategy above is useless. Or is it?
your job is to write a program that determines if a position of S-Nim is a losing or a winning position. A position is a winning position if there is at least one move to a losing position. A position is a losing position if there are no moves to a losing position. This means, as expected, that a position with no legal moves is a losing position.
For each test case: The rst line contains a number k (0 < k <= 100) describing the size of S, followed by k numbers si (0 < si <= 10000) describing S. The second line contains a number m (0 < m <= 100) describing the number of positions to evaluate. The next m lines each contain a number l (0 < l <= 100) describing the number of heaps and l numbers hi (0 <= hi <= 10000) describing the number of beads in the heaps.
The last test case is followed by a 0 on a line of its own.OutputFor each position:
If the described position is a winning position print a 'W'.
If the described position is a losing position print an 'L'.
Print a newline after each test case.Sample Input
2 2 5 3 2 5 12 3 2 4 7 4 2 3 7 12 5 1 2 3 4 5 3 2 5 12 3 2 4 7 4 2 3 7 12 0Sample Output
LWW WWL
题目大意：nim游戏变形，每次每堆石子只能取出恰好给定的数值之一的石子，如石子数还有10，可取数字为（2,5,7,13），那实际上取的方法就只有：每次取2个总共取5次，或每次取5个总共取2次，7和13实际上并没有什么用
思路，将SG函数求出来然后求nim和
SG函数求法
void getSG()
{
memset(SG,0,sizeof SG);
for(int i=1;i<=10005;i++){//每堆石子的石子数最大值为10000，所以循环至10000就好了
memset(vis,0,sizeof vis);//vis数组用于标记当前mex集合里有哪些元素，因为每堆石子能够构成它的元素的集合可能是不同的， //所以每次都需要将vis数组置0
for(int j=1;j<=N;j++)
{
if(num[j]>i)continue;//如果当前石子总数小于想取的石子数num[j]，那就不取，所以continue了
vis[SG[i-num[j]]]=1;//如果能取，那么就是取的这一堆石子数肯定小于i，所以它的SG值肯定已经算过了，由SG函数定义可知 //所取石子数的对应SG值肯定是总石子数为i所对应SG函数的mex集合中的元素，于是将其标记
}
for(int j=0;;j++)if(!vis[j]){//从小到大找不在集合中的最小非负整数
SG[i]=j;
break;//找到就不用找了
}
}
}
AC代码
#include scanf("%d",&n);
int temp=0,tempnum;
for(int j=0;jscanf("%d",&tempnum);
temp^=SG[tempnum];
}
if(temp)printf("W");
else printf("L");
}
printf("\n");
}
return 0;
}